Answer:
a) 17.5
b) 15.6
c) 13.3
d) 21.51
Step-by-step explanation:
The given function is equal to:
f(x)=kx^2
where
[tex]\int\limits^y_0 {kx^{2} } \, =1[/tex]
where y=23
Clearing k=0.00025
a) [tex]Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5[/tex]
b)[tex]Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2} =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6[/tex]
c) The function is equal to:
f(x)=k(1+2x)
[tex]\int\limits^y_0 {k(1+2x)} \, =1[/tex]
where y=20
k=0.0024
[tex]Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3[/tex]
d) [tex]Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2} =198.4-176.89=21.51[/tex]
